Question

(9-x)(6x-1)(x-7)≥0пж пж пж пж пж​

Answer 1

$(9 - x)(6x - 1)(x - 7) = 0$

9-х=0 или 6х-1=0 или х-7=0

х=9 или х=1/6 или х=7

..+.........-.........+........-

----(1/6)----(7)----(9)------->

Ответ: х€(-беск;1/6] и [7;9]

Answer 2

To solve the inequality $(9 - x)(6x - 1)(x - 7) \geq 0$, we need to find the intervals on the number line where the expression is non-negative (greater than or equal to zero).

The expression will be equal to zero at the points where any of the factors is equal to zero. So, we set each factor equal to zero and solve for $x$:

1. $9 - x = 0$: $x = 9$
2. $6x - 1 = 0$: $x = \frac{1}{6}$
3. $x - 7 = 0$: $x = 7$

Now, we'll use these points to divide the number line into intervals:

• Interval 1: $(-\infty, \frac{1}{6})$
• Interval 2: $(\frac{1}{6}, 7)$
• Interval 3: $(7, 9)$
• Interval 4: $(9, \infty)$

Now, we check the sign of the expression $(9 - x)(6x - 1)(x - 7)$ in each interval:

• Interval 1: Pick a test point $x = 0$, and plug it into the expression. $(9 - 0)(6 \cdot 0 - 1)(0 - 7) = 9 \cdot 1 \cdot (-7) < 0$.
• Interval 2: Pick a test point $x = 3$, and plug it into the expression. $(9 - 3)(6 \cdot 3 - 1)(3 - 7) = 6 \cdot 17 \cdot (-4) > 0$.
• Interval 3: Pick a test point $x = 8$, and plug it into the expression. $(9 - 8)(6 \cdot 8 - 1)(8 - 7) = 1 \cdot 47 \cdot 1 > 0$.
• Interval 4: Pick a test point $x = 10$, and plug it into the expression. $(9 - 10)(6 \cdot 10 - 1)(10 - 7) = (-1) \cdot 59 \cdot 3 < 0$.

Therefore, the solution to the inequality is:

$x \in \left( -\infty, \frac{1}{6} \right] \cup (7, 9] \cup (9, \infty)$

Note: Square brackets $[ \, ]$ denote that the endpoint is included, while round brackets $( \, )$ denote that the endpoint is not included.